Astroid

${Substituting}$ ${ x}$ = $a(\cos \theta)^3$ & y = $a(\sin\theta)^3$

${Area}$ = $\int ydx$

${x}$ = $a(\cos\theta)^3$

${dx}$ = $3a(\cos\theta)^2 \ \times-(sin\theta) d\theta$ ${Area }$ = $\int a(\sin\theta)^3 \times -3a(\cos\theta)^2 \times (\sin\theta) d\theta$

${Area }$ = $-3a^2\int (\sin\theta)^3 (\cos\theta)^2 \times (\sin\theta) d\theta$

${ Total}$ ${Area}$ ${ enclosed (A)}$ = $4 \times$ $-3a^2\int_{0}^{\pi/2}( \sin\theta)^4( \cos\theta)^2 d\theta$

$\int_{0}^{\pi/2} (\sin\theta)^4 (\cos\theta)^2 d\theta$ = $\Gamma$ $\frac{4+1}{2}$ $\Gamma$ $\frac{2+1}{2}$ $\times$ $\dfrac {1}{2}$ $\times$ $1/ \Gamma$ $\frac{4+2+2}{2}$

$\int_{0}^{\pi/2} (\sin\theta)^4 (\cos\theta)^2 d\theta$ = $\Gamma$ $\dfrac{5}{2}$ $\Gamma$ $\dfrac{3}{2}$ $\times$ $\dfrac {1}{2}$ $\times$ $\dfrac{1}{\Gamma 4}$

$\int_{0}^{\pi/2} (\sin\theta)^4 (\cos\theta)^2 d\theta$ = $\dfrac { \dfrac{3}{8} \times \pi}{2\times3!}$

${ A}$ = $| -4a^2\times 3 \times$ $\dfrac { \dfrac{3}{8} \times \pi}{2\times3!}|$

${ A}$ = $\dfrac{3 \pi a^2}{8}$

Sorry, for the bad latex coding I did with those Gamma functions.